Question

A mass ${}^{\prime }{m}^{\prime }$ is released, with a horizontal speed $v$ from the top of a smooth and fixed, hemisphere bowl, of radius $r.$ The angle $\theta$ w.r.t. the vertical where it leaves contact with the bowl is

• A. ${\sin }^{-1}(\frac{v^2}{3rg}+\frac{2}{3})$
• B. ${\cos }^{-1}(\frac{v^2}{3rg}+\frac{2}{3})$
• C. ${\tan }^{-1}(\frac{v^2}{3rg}+\frac{2}{3})$
• D. ${\cos }^{-1}\left(2/3\right)$

Answer 1

Answer: (B)

${\cos }^{-1}(\frac{v^2}{3rg}+\frac{2}{3})$

$\begin{array}{c}\text{Given, radius of hemisphere}=r\\ \text{mass of block}=m\\ \text{speed of block (horizontal)}=v\end{array}$

$\begin{array}{c}\text{Let, the highest point be P, where the}\\ \text{block is horizontally projected with}\\ \text{velocity}v\text{.}\\ \text{and, let the block covers vertically a}\\ \text{distance 'h' before leaving the surface.}\end{array}$

$\begin{array}{cc}\text{so,}R=h+R\cos \theta \Rightarrow h=R\left(1\cos \theta \right)& \\ \text{we know,}{\left(V^{\prime }\right)}^2& =v^2+2gh[V^{\prime }\text{is velocity at Q}\text{where it leaves}\\ & =v^2+2gR(1-\cos \theta )\end{array}$

$\begin{array}{c}\text{and,}mg\cos \theta -N=\frac{m{\left(V^{\prime }\right)}^2}{R}\\ \text{since, contact is lost at}Q,N=0\text{,}\\ \therefore mg\cos \theta =\frac{m{\left(V^{\prime }\right)}^2}{R}\Rightarrow Rg\cos \theta =v^2+2gR(1-\cos \theta )\end{array}$

$\therefore [\cos \theta =\frac{v^2}{3gR}+\frac{2}{3}]$

$\text{Hence,}\theta ={\cos }^{-1}(\frac{v^2}{3gR}+\frac{2}{3})\text{(option B is correct).}$